Lecture02

Physics 101: Lecture 2, Pg 1

Forces: Equilibrium Examples Forces: Equilibrium Examples

Physics 101: Lecture 02

Today’s lecture will cover Textbook Sections 2.1-2.7

Phys 101 URL:http://online.physics.uiuc.edu/courses/phys101

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Exam I


OverviewOverview Last Lecture

Newton’s Laws of Motion» Inertia» F=ma» Pairs

Free Body Diagrams» Draw coordinate axis, each direction is independent.» Simple Picture» Identify/draw all forces

Friction: kinetic f = kN; static f ≤ s NGravity W = m g (near Earth’s surface!)

TodayContact Force---SpringsContact Force---Tension2-D Examples


Free Body DiagramsFree Body Diagrams Choose Object (book) Label coordinate axis Identify All Forces

Hand (to right)Gravity (down)Normal (table, up)Friction (table, left)

Physics

hand

Gravity

Normal

friction

y

x


Book Pushed Across TableBook Pushed Across Table Calculate force of hand to keep the book sliding at a

constant speed, if the mass of the book is 1 Kg, s = .84 and k=.75.

Physics

hand

Gravity

Normal

frictiony

x

x-direction: F=0Fhand-Ffriction = 0 Fhand=Ffriction

Fhand=k FNormal

y-direction: F=0FNormal-FGravity = 0FNormal = FGravity

FNormal =x 9.8=9.8 N

Combine: Fhand=k FNormal

Fhand=0.75 x 9.8 N

Fhand=ewtons

Constant Speed => F=0


Contact Force: SpringsContact Force: Springs

Force exerted by a spring is directly proportional to its displacement (stretched or compressed). Fspring = -k x

Example: When a 5 kg mass is suspended from a spring, the spring stretches 8 cm. If it is hung by two identical springs, they will stretch

A) 4 cmB) 8 cm C) 16 cmF1 +F2 - Fgravity = 0

F1 +F2 = Fgravity

k1x1 + k2x2= m g

2 k x = mg

y

x

x = m g / (2k)

= (½ m) g / k

We know mg/k = 8 cm.

So: ½ mg /k = 4 cm.

F2

Fgravity

F1


Contact Force: SpringsContact Force: Springs

Force exerted by a spring is directly proportional to its displacement (stretched or compressed). Fspring = -k x

Example: When a 5 Kg mass is suspended from a spring, the spring stretches 8 cm. If it is hung by two identical springs, they will stretch

A) 4 cmB) 8 cm C) 16 cm

y

x


Contact Force: TensionContact Force: Tension Tension in an Ideal String:

Magnitude of tension is equal everywhere.Direction is parallel to string (only pulls)

Example : Determine force applied to string to suspend 45 kg mass hanging over pulley: Answer: FBDF = ma

F = mg= 440 Newtons


Pulley ACTPulley ACT Two boxes are connected by a string over a

frictionless pulley. In equilibrium, box 2 is lower than box 1. Compare the weight of the two boxes.

A) Box 1 is heavier

B) Box 2 is heavier

C) They have the same weight1

2F = m a1) T - m1 g = 02) T – m2 g = 0=> m1 = m2

1

T

m1g 2

T

m2g


Tension Example: Tension Example: Determine the force exerted by the hand to

suspend the 45 kg mass as shown in the picture.

y

x

F = m aT + T – W = 02 T = WT = m g / 2 = (45 kg x (9.8 m/s2)/ 2 = 220 N

•Remember the magnitude of the tension is the same everywhere along the rope!

T T

W

A) 220 N B) 440 N C) 660 N

D) 880 N E) 1100 N


Tension ACT IITension ACT II Determine the force exerted by the ceiling to

suspend pulley holding the 45 kg mass as shown in the picture.

y

x

F = m a

Fc -T - T – T = 0

Fc = 3 T

Fc = 3 x 220 N = 660 N•Remember the magnitude of the tension is the same everywhere along the rope!

Fc

T

A) 220 N B) 440 N C) 660 N

D) 880 N E) 1100 N


Springs Preflight Springs Preflight •What does scale 1 read? (90% got correct!)

• A) 225 N B) 550 N C) 1100 N

1

The magnitude of tension in a ideal string is equal everywhere.


Springs ACTSprings ACT•Scale 1 reads 550 Newtons. What is the reading on scale 2?

A) 225 N B) 550 N C) 1100 N

In both cases the NET FORCE on the spring is zero, and the force to the right is 550N. Therefore the force to the left is also 550 N.

1 2


Forces in 2 Dimensions: RampForces in 2 Dimensions: Ramp Calculate tension in the rope necessary to

keep the 5 kg block from sliding down a frictionless incline of 20 degrees.

1) Draw FBD

2) Label Axis

y

x

T

N

WNote, weight is not in x or y direction! Need to decompose it!


Forces in 2 Dimensions: RampForces in 2 Dimensions: Ramp Calculate force necessary to keep the 5 kg

block from sliding down a frictionless incline of 20 degrees.

y

x

W

W sin

W cos

T

N

W


Forces in 2 Dimensions: RampForces in 2 Dimensions: Ramp Calculate force necessary to keep the 5 kg

block from sliding down a frictionless incline of 20 degrees.

y

x

x- direction

W sin – T = 0

T = W sinm g sinNewtons

T

N

W


Normal Force ACTNormal Force ACT

What is the normal force of ramp on block?

A) FN > mg B) FN = mg C) FN < mg

T

N

W

In “y” directionF = maN – W cos = 0N = W cos

y

x

N = m g cos

W

W sin

W cos


Force at Angle ExampleForce at Angle Example A person is pushing a 15 kg block across a floor with k= 0.4 at a

constant speed. If she is pushing down at an angle of 25 degrees, what is the magnitude of her force on the block?

y

x

Normal

Weight

Pushing

x- direction: Fx = max

Fpush cos() – Ffriction = 0 Fpush cos() – FNormal = 0 FNormal = Fpush cos() /

y- direction: Fy = may

FNormal –Fweight – FPush sin() = 0 FNormal –mg – FPush sin() = 0

Combine: (Fpush cos() / –mg – FPush sin() = 0 Fpush ( cos() / - sin()) = mg Fpush = m g / ( cos()/ – sin())

Friction

Fpush = 80 N


SummarySummary Contact Force: Spring

Can push or pull, force proportional to displacementF = k x

Contact Force: TensionAlways Pulls, tension equal everywhereForce parallel to string

Two Dimensional ExamplesChoose coordinate systemAnalyze each direction is independent

Technology

Lecture02